Prealgebra (7th Edition)

Published by Pearson
ISBN 10: 0321955048
ISBN 13: 978-0-32195-504-3

Appendix B - Exercise Set: 8

Answer

$\frac{1}{3}a^3b^5$

Work Step by Step

$\frac{9a^4b^7}{27ab^2}=(\frac{9}{27})(\frac{a^4}{a})(\frac{b^7}{b^2})=(\frac{/\!\!9}{/\!\!9\times3})(a^{4-1})(b^{7-2})=\frac{1}{3}a^3b^5$
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